Question

A particle executes SHM with time period $T$ and amplitude $A$.The maximum possible average velocity in time $\frac{T}{4}$ is

• A. $\frac{2A}{T}$
• B. $\frac{4A}{T}$
• C. $\frac{8A}{T}$
• D. $4\sqrt{2}\frac{A}{T}$

Answer 1

The correct option is D $4\sqrt{2}\frac{A}{T}$

Maximum average velocity magnitude for $\frac{T}{4}$ time will occur in the time from $\frac{-T}{8}$ to $0$ to $\frac{+T}{8}$,
we can Integrate $A\omega \cos \omega tdt$ within the limits from $\frac{-T}{8}$ to $\frac{+T}{8}$, and divide the answer by $\frac{T}{4}$
here $\omega =\frac{2\pi }{T}$, that gives, the maximum average velocity $4\sqrt{2}\frac{A}{T}$.